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Applications of LCM & HCF
•   Greatest Number that will leave no remainder when divides a, b and c

    Required number = HCF of a, b and c


Example
    A shopkeeper has three cakes of weight 18 kg, 45 kg and 36 kg. If he
    wants to make these cakes into pieces of equal weight without
    wastage, what is the maximum possible weight of each piece?

    (1) 1             (2) 9           (3) 12          (4) 13
Applications of LCM & HCF
•       The greatest number that will divide a, b & c leaving
    remainder x, y & z respectively

         Required number = HCF of {(a – x), (b – y), (c – z)}

Example
         What is the greatest number that divide 20, 50, and 40
    leaving 2, 5 and 4 as remainder?
         Here             a–x=
                          b–y=
                          c–z=
         Required Number =
Applications of LCM & HCF
• To find the greatest number that will divide x, y and z leaving the
  same remainder “r” in each case.

   Required number = HCF of (x – r), (y – r) and (z – r)
Example
        Find the greatest number which will divide 369, 449, 689,
        5009 and 729 so as to leave the remainder 9 in each case
        (A) 2          (B) 49           (C) 35          (D) 40

        Required Number is
        HCF of {(369-9), (449-9), (689-9), (5009 – 9), (729 – 9)}
               = HCF of {360, 440, 680, 5000, 720}
               = 40
Applications of LCM & HCF
•        To find the greatest number that will divide x, y and z
    leaving the same remainder in each case.
         Required number = HCF of |x – y|, |y – z| and |z – x|

Example
         What is the greatest number that will divide 1305, 4665 and
         6905 leaving in each case the same remainder?           Also
         calculate the remainder.
         (1) 1210, 158                   (2) 1120, 158
         (3) 1120, 185                   (4) 1210, 185
Solution
     Here we have
       |4665 – 1305|, |6905 – 4665|, |6905 – 1305|
     =      3360,         2240,          5600
     =    1120 x 3,      1120 x 2,      1120 x 5
     Required number = HCF of (3360, 2240, 5600) = 1120

     To calculate Remainder
     1305 = 1120 x 1 + 185
     So the Remainder = 185
Applications of LCM & HCF
•   Least number which is exactly divisible by a, b, c
    Required number = LCM of x, y and z
    Sequence           = n x LCM
                         (Where n is a natural Number)
Example
    Find the greatest number of five digits which is divisible by
    32, 36, 40, 42 and 48
    (a) 999720      (b) 90702        (C) 90720         (d) 90730
Solution

Step I
           LCM of (32, 36, 40, 42, 48) = 10080
Step II
           Find greater number of 5 digit which multiple of 10080


                 10080 99999        9
                       90720
                        9279
    Greatest number = 90720
Applications of LCM & HCF
•    To find least number which when divided by a, b, c leaves
    “r” as a remainder
    Required number = LCM of (a, b, c) + r

Example
    What is the smallest sum which a person can have such that
    when he distributed @ Rs. 2.5 or Rs 20 or Rs 12 or Rs. 7.5 per
    person in a group, he is always left with Rs. 2.00?
    (A) Rs. 62         (B) Rs. 80        (C) Rs. 90     (D) Rs. 100
Solution
     According to the problem the person must have the money
     equal to the LCM of 2.5, 20, 12 and 7.5 and the remainder
     money always left.
Applications of LCM & HCF
•      Least number which when divided by a, b, c leaves x, y, z
       as remainder, such that a – x = b – y = c – z = k (say)
       Required number = LCM of (a, b, c) – k

Example
       What is the greatest number of 4 digits that when divided
       by the numbers 6, 9, 12, 17 leaves 5, 8, 11, 16 as
    remainders respectively?
       (a) 9791        (b) 9793        (c) 9895         (d) 9497
Solution

Step 1 Here,
       6 – 5 = 9 – 8 = 12 – 11= 17 – 16 = 1 = K (let)
Step 2 LCM of 6, 9, 12, 17 = 17 x 9 x 4 = 17 x 36 = 612
Step 3 Find Greatest 4 digit number which is multiple of 612

                   612    9999
                           612        16

                           3879
                                           Greatest Number
                           3672
                                           = 9999 – 207 = 9792
                            207
 Step 4   Required number = (LCM) n - K
                          = 9792 - 1         = 9791
Thank you !!

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Ppt On Lcm & Hcf Questions For Cat Preparation

  • 1. Applications of LCM & HCF • Greatest Number that will leave no remainder when divides a, b and c Required number = HCF of a, b and c Example A shopkeeper has three cakes of weight 18 kg, 45 kg and 36 kg. If he wants to make these cakes into pieces of equal weight without wastage, what is the maximum possible weight of each piece? (1) 1 (2) 9 (3) 12 (4) 13
  • 2. Applications of LCM & HCF • The greatest number that will divide a, b & c leaving remainder x, y & z respectively Required number = HCF of {(a – x), (b – y), (c – z)} Example What is the greatest number that divide 20, 50, and 40 leaving 2, 5 and 4 as remainder? Here a–x= b–y= c–z= Required Number =
  • 3. Applications of LCM & HCF • To find the greatest number that will divide x, y and z leaving the same remainder “r” in each case. Required number = HCF of (x – r), (y – r) and (z – r) Example Find the greatest number which will divide 369, 449, 689, 5009 and 729 so as to leave the remainder 9 in each case (A) 2 (B) 49 (C) 35 (D) 40 Required Number is HCF of {(369-9), (449-9), (689-9), (5009 – 9), (729 – 9)} = HCF of {360, 440, 680, 5000, 720} = 40
  • 4. Applications of LCM & HCF • To find the greatest number that will divide x, y and z leaving the same remainder in each case. Required number = HCF of |x – y|, |y – z| and |z – x| Example What is the greatest number that will divide 1305, 4665 and 6905 leaving in each case the same remainder? Also calculate the remainder. (1) 1210, 158 (2) 1120, 158 (3) 1120, 185 (4) 1210, 185
  • 5. Solution Here we have |4665 – 1305|, |6905 – 4665|, |6905 – 1305| = 3360, 2240, 5600 = 1120 x 3, 1120 x 2, 1120 x 5 Required number = HCF of (3360, 2240, 5600) = 1120 To calculate Remainder 1305 = 1120 x 1 + 185 So the Remainder = 185
  • 6. Applications of LCM & HCF • Least number which is exactly divisible by a, b, c Required number = LCM of x, y and z Sequence = n x LCM (Where n is a natural Number) Example Find the greatest number of five digits which is divisible by 32, 36, 40, 42 and 48 (a) 999720 (b) 90702 (C) 90720 (d) 90730
  • 7. Solution Step I LCM of (32, 36, 40, 42, 48) = 10080 Step II Find greater number of 5 digit which multiple of 10080 10080 99999 9 90720 9279 Greatest number = 90720
  • 8. Applications of LCM & HCF • To find least number which when divided by a, b, c leaves “r” as a remainder Required number = LCM of (a, b, c) + r Example What is the smallest sum which a person can have such that when he distributed @ Rs. 2.5 or Rs 20 or Rs 12 or Rs. 7.5 per person in a group, he is always left with Rs. 2.00? (A) Rs. 62 (B) Rs. 80 (C) Rs. 90 (D) Rs. 100
  • 9. Solution According to the problem the person must have the money equal to the LCM of 2.5, 20, 12 and 7.5 and the remainder money always left.
  • 10. Applications of LCM & HCF • Least number which when divided by a, b, c leaves x, y, z as remainder, such that a – x = b – y = c – z = k (say) Required number = LCM of (a, b, c) – k Example What is the greatest number of 4 digits that when divided by the numbers 6, 9, 12, 17 leaves 5, 8, 11, 16 as remainders respectively? (a) 9791 (b) 9793 (c) 9895 (d) 9497
  • 11. Solution Step 1 Here, 6 – 5 = 9 – 8 = 12 – 11= 17 – 16 = 1 = K (let) Step 2 LCM of 6, 9, 12, 17 = 17 x 9 x 4 = 17 x 36 = 612 Step 3 Find Greatest 4 digit number which is multiple of 612 612 9999 612 16 3879 Greatest Number 3672 = 9999 – 207 = 9792 207 Step 4 Required number = (LCM) n - K = 9792 - 1 = 9791